$\text{Na}_2\text{S}_2\text{O}_3$ is reacted with dilute $\text{HCl}$ and a pale yellow precipitate is produced. When $1\,\text{cm}^3$ of $0.1\,\text{mol dm}^{-3}$ $\text{HCl}$ is mixed with $10\,\text{cm}^3$ of $0.02\,\text{mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$, the precipitate appears slowly. When the test is carried out again using $1\,\text{cm}^3$ of $0.1\,\text{mol dm}^{-3}$ $\text{HCl}$ together with $10\,\text{cm}^3$ of $0.05\,\text{mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$, the precipitate appears at a faster rate. What accounts for this?
- AThe activation energy of the reaction is lower when $0.05\,\text{mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$ is used.
- BThe collisions between reactant particles are more violent when $0.05\,\text{mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$ is used.
- CThe reactant particles collide more frequently when $0.05\,\text{mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$ is used.
- DThe reaction proceeds by a different pathway when $0.05\,\text{mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$ is used.